223_pdfsam_math 54 differential equation solutions odd

# 223_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 4.6 From the ﬁrst equation in the above system we also obtain u1 (t) = −u2 (t) tan 2t = − Integrating yields u1 (t) = − u2 (t) = Thus, vp (t) = − 1 1 sec3 2t cos 2t + (sec 2t tan 2t + ln | sec 2t + tan 2t|) sin 2t 12 8 1 1 1 2 2 = − sec 2t + tan 2t + sin 2t ln | sec 2t + tan 2t| 12 8 8 11 1 2 sec 2t − + sin 2t ln | sec 2t + tan 2t|, = 24 88 1 11 sec2 2t − + sin 2t ln | sec 2t + tan 2t|. 24 88 1 2 1 2 sec4 2t sin 2t dt = − sec3 2t dt = 1 2 cos−4 2t sin 2t dt = − 1 sec3 2t, 12 1 sec4 2t sin 2t . 2 1 (sec 2t tan 2t + ln | sec 2t + tan 2t|). 8 and a general solution to the given equation is v (t) = c1 cos 2t + c2 sin 2t + 15. The corresponding homogeneous equation is y + y = 0. Its auxiliary equation has the roots r = ±i. Hence, a general solution to the homogeneous problem is given by yh (t) = c1 cos t + c2 sin t. We will ﬁnd a particular solution to the original equation by ﬁrst ﬁnding a particular solution for each of two problems, one with the nonhomogeneous term g1 (t) = 3 sec t and the other one with the nonhomogeneous term g2 (t) = −t2 + 1. Then we will use the superposition principle to obtain a particular solution for the original equation. The term 3 sec t is not in a form that allows us to use the method of undetermined coeﬃcients. Therefore, we will use the method of variation of parameters. To this end, let y1 (t) = cos t and y2 (t) = sin t (linearly independent solutions to the corresponding homogeneous problem). Then a particular solution yp,1 to y + y = 3 sec t has the form yp,1(t) = v1 (t)y1 (t) + v2 (t)y2 (t) = v1 (t) cos t + v2 (t) sin t, 219 ...
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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